question_answer

If \[\left| {{x}^{2}}-2x+2 \right|-\left| 2{{x}^{2}}-5x+2 \right|\]=\[\left| {{x}^{2}}-3x \right|\], then the set of values of x is

A) \[\left( -\infty ,0 \right]\cup \left[ 3,\infty  \right)\]

B) \[\left[ 0,\frac{1}{2} \right]\cup \left[ 2,3 \right]\]

C) \[\left( -\infty ,0 \right]\cup \left[ \frac{1}{2},2 \right]\cup \left[ 3,\infty  \right)\]

D) \[\left[ 0,2 \right]\cup \left[ 3,\infty  \right)\]

Correct Answer: B

Solution :

[b] we have,             \[\left| 2{{x}^{2}}-5x+2 \right|+\left| {{x}^{2}}-3x \right|=\left| {{x}^{2}}-2x+2 \right|\] =\[\left| 2{{x}^{2}}-5x+2)-({{x}^{2}}-3x) \right|\] \[\therefore (2{{x}^{2}}-5x+2)({{x}^{2}}-3x)\le 0\] \[\Rightarrow (2x-1)(x-2)x(x-3)\le 0\] From the sign scheme, we have \[x\in \left[ 0,\frac{1}{2} \right]\cup [2,3]\]